Question

The locus of a point equidistant from two points whose position vectors are $\vec{a}$ and $\vec{b}$, is

• A. $\left\{\vec{r}-\frac{1}{2}\left(\vec{a}+\vec{b}\right)\right\}\cdot \left(\vec{b}-\vec{a}\right)=0$
• B. $\left\{\vec{r}-\left(\vec{a}+\vec{b}\right)\right\}\cdot \vec{b}=0$
• C. $\left\{\vec{r}-\frac{1}{2}\left(\vec{a}+\vec{b}\right)\right\}\cdot \vec{a}=0$
• D. $\left\{\vec{r}-\frac{1}{2}\left(\vec{a}+\vec{b}\right)\right\}\left(\vec{a}+\vec{b}\right)=0$

Answer 1

Answer: (A)

Locus of the point $\vec{r}$ equidistant from two points is given by
${\left|\vec{r}-\vec{a}\right|}^2={\left|\vec{r}-\vec{b}\right|}^2$
$\Rightarrow {\left|\vec{r}\right|}^2+{\left|\vec{a}\right|}^2-2\left(\vec{r}\cdot \vec{a}\right)={\left|\vec{r}\right|}^2+{\left|\vec{b}\right|}^2-2\left(\vec{r}\cdot \vec{b}\right)$
$\Rightarrow 2\vec{r}\cdot \left(\vec{b}-\vec{a}\right)={\left|\vec{b}\right|}^2-{\left|\vec{a}\right|}^2$
$\Rightarrow \vec{r}\cdot \left(\vec{b}-\vec{a}\right)=\frac{1}{2}\left\{\left(\vec{b}+\vec{a}\right)\cdot \left(\vec{b}-\vec{a}\right)\right\}$
$\Rightarrow \left\{\vec{r}-\frac{1}{2}\left(\vec{b}+\vec{a}\right)\right\}.\left(\vec{b}-\vec{a}\right)=0$